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basic penetrating Q (5e)


secretID

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1d6 RKA, Penetrating vs. 15 Resistant DEF, not Hardened. A "3" is rolled, with a STUN multiplier of 3. The target takes 1 BODY, but does the target take 0 or 3 STUN? I think it's 0.

 

This from the FAQ wasn't completely clear, but seems to indicated it would be 0:

 

"If a character buys Penetrating for a Killing Attack, does it affect the STUN Multiplier or the amount of STUN damage done to/taken by the target?

 

Ordinarily, no — roll the standard STUN Multiplier and apply the STUN damage using the normal rules for Killing Damage. However, a character could buy Penetrating for a KA and declare that it applies to the STUN damage, not the BODY damage. Presumptively, Penetrating for a KA applies to BODY damage; the character must specifically indicate to the GM that he’s buying it for a KA’s STUN instead."

 

Thanks.

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Re: basic penetrating Q (5e)

 

1d6 RKA, Penetrating vs. 15 Resistant DEF, not Hardened. A "3" is rolled, with a STUN multiplier of 3. The target takes 1 BODY, but does the target take 0 or 3 STUN? I think it's 0.

 

This from the FAQ wasn't completely clear, but seems to indicated it would be 0:

 

"If a character buys Penetrating for a Killing Attack, does it affect the STUN Multiplier or the amount of STUN damage done to/taken by the target?

 

Ordinarily, no — roll the standard STUN Multiplier and apply the STUN damage using the normal rules for Killing Damage. However, a character could buy Penetrating for a KA and declare that it applies to the STUN damage, not the BODY damage. Presumptively, Penetrating for a KA applies to BODY damage; the character must specifically indicate to the GM that he’s buying it for a KA’s STUN instead."

 

Thanks.

 

Except that you always have to take as many Stun as you take Body. Unless I'm suffering from Versionitus...

 

Edit: Checked, nope, still there in 6e. So the target would take 1 Body and 1 Stun.

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